The equation of a circle $C$ is $x^2+y^2-2x-2y-34 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2-2x) + (y^2-2y) = 34$ $(x^2-2x+1) + (y^2-2y+1) = 34 + 1 + 1$ $(x-1)^{2} + (y-1)^{2} = 36 = 6^2$ Thus, $(h, k) = (1, 1)$ and $r = 6$.